php - <td> 不填充所有动态表 HTML PHP MySQL

Question

我的表没有填满我需要的所有数据。缺少一些列被填充表格多个。 th 是 Renault 的列需要相同，因为所有其余的都是空的。

在下一张图片中，您可以看到当我在动态选择框中选择多个选项并提交时，仅填充第一列，而预期的是填充所有选定的列。

接下来是给选择框的值倍数的代码。

if ($consulta !=0) {
}//
else{
    echo "";
}

@$s1= $_POST['s1'];
if( is_array($s1)){
    while (list ($key, $val) = each ($s1)) {

    }
}//
else{
    echo "";
}
/////////////////
@$s2= $_POST['s2'];
if( is_array($s2)){

?>

接下来是在html表格中填充数据的代码。那行代码<td align=\"center\" bgcolor='FFFFFF'>需要关联以下结果:<?php echo "$key => $val1\n";?>以何种形式:

<?php echo "$key => $val1\n";?> 的结果

0 => Renault 1 => Opel 2 => Mercedes 3 => Audi 

缺少预期的输出。

雷诺

<form onsubmit=\"return validate();\" id=\"teste\" method=\"post\" action=\"teste.php\">

欧宝

<form onsubmit=\"return validate();\" id=\"teste\" method=\"post\" action=\"teste.php\">

奔驰

<form onsubmit=\"return validate();\" id=\"teste\" method=\"post\" action=\"teste.php\">

奥迪

<form onsubmit=\"return validate();\" id=\"teste\" method=\"post\" action=\"teste.php\">

这是动态的，将从 mysql 数据库中获取值，我可以选择我想要的。

所有的表格都填满了数据。静态和动态。

    <table>
     <tbody>
        <tr>
            <th>Parametro:</th>

    <?php 
        while (list($key, $val1) = each ($s2)) { 
    ?>

            <th><?= htmlspecialchars($val1, ENT_HTML5 | ENT_COMPAT, 'UTF-8') ?></th>
    <?php 
            echo "$key => $val1\n";
    ?>
    <?php 
        } 
    ?>

Parametro 的值:

    {
        echo("<tr>
        <td align=\"center\" bgcolor='FFFFFF'>$utilizador</td>

     <td align=\"center\" bgcolor='FFFFFF'>

    <form onsubmit=\"return validate();\" id=\"teste\" method=\"post\" action=\"teste.php\">


    ");

表单中的代码:

$con = mysqli_connect( "localhost","root","","teste" ); // ◄■ CONNECT TO DATABASE.
$dat_menuid = mysqli_query( $con,"SELECT * FROM menu WHERE menuId = '".$menuId."'"  ) // ◄■ MENUIDs.
                      or die( mysqli_error($con) );
        $con1 = mysqli_connect( "localhost","root","","teste" ); // ◄■ CONNECT TO DATABASE.           
            $idtem = mysqli_query($con1, "SELECT MAX(id_tem) as id_tem FROM tem")        
                       or die( mysqli_error($con) );


// Print out result
while($row = $idtem->fetch_assoc())
   $toEcho = $row["id_tem"];

//echo $toEcho;

$toEchosave = $toEcho + 1;
$_SESSION['myvar'] = $toEchosave;




$options = array( "0.00","0.05","0.10","0.15","0.20","0.25","0.30","0.35","0.40","0.45","0.50","0.55","0.60","0.65","0.70","0.75","0.80","0.85","0.90","0.95","1" ); // ◄■■ OPTIONS ARE STATIC (ALWAYS THE SAME).
while ( $row_menuid = mysqli_fetch_array( $dat_menuid ) ) // ◄■ DISPLAY <SELECT>s.


{ echo "<select class='sumselect' name='corp_resp&{$row_menuid['menuId']}&{$_SESSION['UtilizadorID']}&{$dateTime}&{$toEchosave}' multiple>"; // ◄■■ CORP_RESP&1,CORP_RESP&2.

    foreach ($options as $opt)
                echo "<option value='$opt'>$opt</option>\n";
            echo "</select>\n";
        } ?>
    </form>
    </td>

  </tr>
  <?php } ?>


 <?php }
   }
   }

   ?>
 <div id='sum'></div>
 <br><br><br>


</tbody>
</table>

Answer 1

遵循此代码

<?php
$sql = "SELECT * FROM menu WHERE $id = menuIdPai  ORDER BY menuId ";

$consulta = mysql_query($sql);

    while ($mostrar = mysql_fetch_array($consulta)) {

    $id = $mostrar['menuId'];
    $utilizador = $mostrar['menuNome'];
    ?>

    <tr>
        <td align=\"center\" bgcolor='FFFFFF'>Test1</td>

        <td align=\"center\" bgcolor='FFFFFF'>Test2</td>
        <td align=\"center\" bgcolor='FFFFFF'>Test3</td>

        <td align=\"center\" bgcolor='FFFFFF'>Test4</td>
    </tr>
    <?php
}

?>

这是主要格式，如果可以，请更新您的代码。你添加 2 td 但你的主要 th 4 所以需要 4 td

你的这个标签没有关闭

 <td align="center" bgcolor='FFFFFF'>

<form onsubmit="return validate();" id="teste" method="post" action="teste.php">

所以在你的脚本中关闭它

    <div id='sum'></div>
   <br><br><br>


 </form>
</td>

 </tr>

将这段代码移到结束循环之前

      foreach ($options as $opt)
                echo "<option value='$opt'>$opt</option>\n";
            echo "</select>\n";
        } ?>
    </form>
    </td>

  </tr>
  <?php } ?>


 <?php }
   }
   }

   ?>
 <div id='sum'></div>
 <br><br><br>


</tbody>
</table>